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\(a,\)\(\frac{2}{\sqrt{x^2-x+1}}\)
\(đkxđ\Leftrightarrow\hept{\begin{cases}x^2-x+1\ge0\\x^2-x+1\ne0\end{cases}\Rightarrow x^2-x+1>0}\)
Mà \(x^2-x+1=x^2-2.\frac{1}{2}x+\frac{1}{4}+\frac{3}{4}\)
\(=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\)với \(\forall x\)
\(\Rightarrow\)Biểu thức luôn được xác định với mọi x
a/ \(x^2-4x+1=\left(x-2+\sqrt{3}\right)\left(x-2-\sqrt{3}\right)\)
Để biểu thức có nghĩa
\(\Leftrightarrow\left(x-2+\sqrt{3}\right)\left(x-2-\sqrt{3}\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge2+\sqrt{3}\\x\le2-\sqrt{3}\end{matrix}\right.\)
b/ Để biểu thức có nghĩa
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{2x-1}{x+3}\ge0\\x+3\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-1\ge0\\x+3\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-1\le0\\x+3\le0\end{matrix}\right.\end{matrix}\right.\\x\ne-3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge\frac{1}{2}\\x\le-3\end{matrix}\right.\\x\ne-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge\frac{1}{2}\\x< -3\end{matrix}\right.\)
\(a,\)\(\frac{1}{1-\sqrt{x^2-3}}\)
\(đkxđ\Leftrightarrow\orbr{\begin{cases}x^2-3\ge0\\x^2-3\ne1\end{cases}}\).
\(x^2-3\ne1\)\(\Rightarrow x^2\ne4\)\(\Rightarrow x\ne\pm2\)
\(x^2-3\ge0\)\(\Rightarrow\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)\ge0\)
Chia trường hợp ra làm nốt nhé
....
\(b,\)\(\frac{x-1}{2-\sqrt{3x+1}}\)
\(đkxđ\Leftrightarrow\orbr{\begin{cases}3x+1\ge0\\\sqrt{3x+1}\ne2\end{cases}}\)
\(3x+1\ge0\)\(\Rightarrow3x\ge-1\)
\(\Rightarrow x\ge\frac{-1}{3}\)
\(\sqrt{3x+1}\ne2\)\(\Rightarrow|3x+1|\ne4\)\(\Rightarrow\hept{\begin{cases}3x-1\ne4\\3x-1\ne-4\end{cases}\Rightarrow\hept{\begin{cases}3x\ne5\\3x\ne-3\end{cases}\Rightarrow}\hept{\begin{cases}x\ne\frac{5}{3}\\x\ne-1\end{cases}}}\)
\(\Rightarrow x\ge-\frac{1}{3}\)và \(x\ne\frac{5}{3}\)
1) \(\frac{1}{\sqrt{2x-1}}\)có nghĩa khi \(\hept{\begin{cases}2x-1\ge0\\\sqrt{2x-1}\ne0\end{cases}}\)
\(\Leftrightarrow2x-1>0\)
\(\Leftrightarrow x>\frac{1}{2}\)
\(\sqrt{5-x}\)có nghĩa khi \(5-x\ge0\Leftrightarrow x\ge5\)
Vậy \(ĐKXĐ:\frac{1}{2}>x\ge5\)
2) \(\sqrt{x-\frac{1}{x}}\)có nghĩa khi \(\hept{\begin{cases}x-\frac{1}{x}\ge0\\x>0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\frac{x^2}{x}-\frac{1}{x}\ge0\\x>0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\frac{x^2-1}{x}\ge0\\x>0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x^2-1\ge0\\x>0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x^2\ge1\\x>0\end{cases}}\)
Vậy \(ĐKXĐ:x\ge1\)
3) \(\sqrt{2x-1}\)có nghĩa khi \(2x-1\ge0\) \(\Leftrightarrow x\ge\frac{1}{2}\)
\(\sqrt{4-x^2}\)có nghĩa khi \(4-x^2\ge0\Leftrightarrow x^2\le4\Leftrightarrow x\le2\)
Vậy \(ĐKXĐ:\frac{1}{2}\le x\le2\)
4) \(\sqrt{x^2-1}\)có nghĩa khi \(x^2-1\ge0\Leftrightarrow x^2\ge1\Leftrightarrow x\ge1\)
\(\sqrt{9-x^2}\)có nghĩa khi \(9-x^2\ge0\Leftrightarrow x^2\le9\Leftrightarrow x\le3\)
Vậy \(ĐKXĐ:1\le x\le3\)
\(a,\)\(\sqrt{\frac{1}{\left(x-3\right)^2}}\)
\(đk:\)\(\frac{1}{\left(x-3\right)^3}\ne0\)\(\Rightarrow\left(x-3\right)^3\ne0\)\(\Leftrightarrow x\ne3\)
Và \(\frac{1}{\left(x-3\right)}>0\Rightarrow x-3>0\)\(\Rightarrow x>3\)
Vậy để căn thức xác định thì x > 3
\(\sqrt{8x-x^2-15}\)
\(=\sqrt{-\left(x^2-8x+15\right)}\)
\(=\sqrt{-\left(x^2-8x+16-1\right)}\)
\(=\sqrt{-\left[\left(x^2-8x+16\right)-1\right]}\)
\(=\sqrt{-\left(x-4\right)^2+1}\)
\(đk:\)\(-\left(x-4\right)^2+1\ge0\)
\(\Rightarrow\left(x-4\right)^2\le1\)
\(\Rightarrow\orbr{\begin{cases}\left(x-4\right)^2=1\\\left(x-4\right)^2=0\end{cases}}\)
\(\left(x-4\right)^2=1\Rightarrow\orbr{\begin{cases}x=5\\x=3\end{cases}}\)
\(\left(x-4\right)^2=0\Rightarrow x=4\)
Vậy căn thức xác định \(\Leftrightarrow x=\left\{3;4;5\right\}\)
Lời giải:
a) ĐKXĐ: \(x^2-x+1>0\)
\(\Leftrightarrow (x-\frac{1}{2})^2+\frac{3}{4}>0\)
\(\Leftrightarrow x\in\mathbb{R}\)
b)
ĐKXĐ: \(\left\{\begin{matrix} x-\sqrt{2x-1}>0\\ 2x-1\geq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} \frac{(2x-1)-2\sqrt{2x-1}+1}{2}>0\\ 2x-1\geq 0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} \frac{(\sqrt{2x-1}-1)^2}{2}>0\\ 2x-1\geq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} \sqrt{2x-1}\neq 1\\ 2x-1\geq 0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\neq 1\\ x\geq \frac{1}{2}\end{matrix}\right.\)
a) ĐKXĐ: \(5x-7\ge0\) \(\Leftrightarrow\)\(x\ge\frac{7}{5}\)
b) ĐKXĐ: \(2x^2+x\ge0\)\(\Leftrightarrow\) \(x\left(2x+1\right)\ge0\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x\ge0\\x\le-\frac{1}{2}\end{cases}}\)
c) ĐKXĐ: \(4-7x\ge0\)\(\Leftrightarrow\)\(x\le\frac{4}{7}\)
d) ĐKXĐ: \(x^3+x\ge0\) \(\Leftrightarrow\)\(x\left(x^2+1\right)\ge0\)\(\Leftrightarrow\)\(x\ge0\)
e) ĐKXĐ: \(\frac{x-5}{2x+1}\ge0\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x\ge5\\x< -\frac{1}{2}\end{cases}}\)
f) ĐKXĐ: \(\frac{3-2x}{3x-2}\ge0\) \(\Leftrightarrow\)\(\frac{2}{3}< x\le\frac{3}{2}\)
\(b,\sqrt{\frac{2x-1}{x+3}}\)
\(Đk:\)\(x+3\ne0\Rightarrow x\ne-3\)
Và \(\frac{2x-1}{x+3}\ge0\)
Khi \(\frac{2x-1}{x+3}=0\Rightarrow2x-1=0\)
\(\Rightarrow2x=1\Rightarrow x=\frac{1}{2}\)
Khi \(\frac{2x-1}{x+3}>0\)\(\Rightarrow\orbr{\begin{cases}2x-1>0;x+3>0\\2x-1< 0;x+3< 0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x>\frac{1}{2};x>-3\\x< \frac{1}{2};x< -3\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x>\frac{1}{2}\\x< -3\end{cases}}\)
Vậy căn thức xác định khi \(x\ge\frac{1}{2};x< -3\)